JEE Advanced · Physics · 9. Gravitation
Gravitational acceleration on the surface of a planet is \(\frac{\sqrt{6}}{11} \mathrm{~g}\), where \(g\) is the gravitational acceleration on the surface of the earth. The average mass density of the planet is \(\frac{2}{3}\) times that of the earth. If the escape speed on the surface of the earth is taken to be 11 \(\mathrm{kms}^{-1}\), the escape speed on the surface of the planet in \(\mathrm{kms}^{-1}\) will be
- A 3
- B 5
- C 7
- D 9
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\(g=\frac{G M}{R^2}=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho}{R^2}\)
or \(\quad g \propto \rho R\)
or \(\quad R \propto \frac{g}{\rho}\)
Now escape velocity, \(v_e=\sqrt{2 g R}\)
or \(\quad v_e \propto \sqrt{g R}\)
or \(\quad v_e \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^2}{\rho}}\)
\(\therefore\left(v_e\right)_{\text {planet }}=\left(11 \mathrm{~km}-\mathrm{s}^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}}\)
\(\therefore\) The correct answer is 3 .
or \(\quad g \propto \rho R\)
or \(\quad R \propto \frac{g}{\rho}\)
Now escape velocity, \(v_e=\sqrt{2 g R}\)
or \(\quad v_e \propto \sqrt{g R}\)
or \(\quad v_e \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^2}{\rho}}\)
\(\therefore\left(v_e\right)_{\text {planet }}=\left(11 \mathrm{~km}-\mathrm{s}^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}}\)
\(\therefore\) The correct answer is 3 .
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